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比较排序的最少比较次数

John Waken posted @ Dec 31, 2009 09:03:01 PM in Algorithm with tags sort , 11060 阅读

比较排序研究

今天在BYR论坛算法版看到一个关于比较排序的问题。题目很经典,如下:给你n个互不相同的数,最少用多少次比较可以完全确定各个数的大小关系。也就是问基于比较的排序最少比较次数是多少?

网上大多数人都是用信息论的思想去理解。这种想法是很好的,不过完全可以正宗的离散数学方法来解决这个问题。下面我给出结果以及证明。

a_{n} 表示将n个互不相同的数(其实相同也没关系,这里是为了让读者更容易理解)排序需要的最少比较次数。

设这n个数是   $$b_{1}, b_{2}, ... \quad,b_{i}, ...\quad b_{n}$$

我们经过m次比较操作确定了每个数之间的关系,记这m次操作如下: $$c_{1},c_{2},...\quad ,c_{i},...\quad c_{m}$$

因为比较操作都是独立的,并不互相影响,所以改变它们的次序对结果无任何影响。我们把所有牵涉到$$b_{n}$$的比较操作都往后移,那么前面的与b_{n}无关的比较操作肯定确定了b_{1}, b_{2}, ... , b_{n-1} 的大小关系,也就是前n-1个数已经排好顺序了。现在只要确定b_{n}在这个排列中的位置就可以了,肯定是用二分查找最快。所以我们可以得到下面的第推式(a字母怎么有两种写法,哪位朋友知道怎么解决吗?):

            $$a_{n} =  a_{n-1} + p_{n-1},  n\geq 3$$                   变换一下得到

            $$a_{n+1} =  a_{n} + p_{n} \quad\quad n \geq 2$$

p_{n}表示往一个含有n个元素的有序排列中插入一个新的元素,并且保证排列仍然有序,所需要的最少比较次数。很多人会认为是 $$log_{2}n$$, 这显然是错误的,因为它可能会得出小数。所以我们要用取整函数要进行调整,我试了下,发现上取整函数、下取整函数都不行,所以我们要发明一个新的取整函 数,我称之为 SuperCeil(超级上取整函数),它和ceil(上取整函数)有点儿像,比如 SuperCeil(1.23) = 2,SuperCeil(2.345) = 3,SuperCeil(7.00001) = 8。但是对整数的处理不一样,ceil(3) = 3,而 SuperCeil(3) = 4;SuperCeil(5) = 6。我想大家应该明白了SuperCeil的意义。这样我们就得到了最终的结果:

           a_{n+1} =  a_{n} + SuperCeil(n) \quad\quad n \geq 2

           初始条件是 a_{2} = 1 , 因为两个数显然比较一次就可以了。


OK,问题应该算解决了,但是还不太严密。关于如何求这个第推式的通项,我还没想出来。先放在此博客上如果您有方法求解,或者发现了错误,请在此博客留言或用邮件通知我。

Chitano 说:
Dec 24, 2010 01:45:32 AM

是说 $$ a_{n+1} = a_{n} + SuperCeil(log_{2}n) \quad\quad n \geq 2$$ 吗? 然而$$a_{4}=5, a_{5}=7=a_{4}+log_{2}4$$ 这里似乎不是SuperCeil吧 另外,过程似乎不太严密,为何前面与$$b_{n}$$无关的操作一定确定了前面n-1个数的顺序?这n-1个数中某些顺序完全可以是在后面与$$b_{n}$$有关的操作中确定的。这里需要证明,一定存在i,使得$$b_{i}$$以外的n-1个数的顺序可以通过与$$b_{i}$$无关的操作的结果决定 另外我可以给出一个下界: 由于n个数的全排列是n!种,而每次比较得到的结果只有2种,因此$$a_{n}$$次比较最多得到$$2^a_{n}$$种可能结果,而每种可能结果最多对应一个n个数的顺序

所以最少比较次数不小于 $$ceil(log_{2}n!)$$

但是我暂时没想到方法证明一定存在这样的比较方案。不过前面到n=8的时候都还正确。

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